Respuesta :
Answer:
B. 0.024
The p-value of the test is 0.024 < 0.05(standard significance level), which means that there is enough evidence that the proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced.
Step-by-step explanation:
Before testing the hypothesis, we need to understand the central limit theorem and subtraction of normal variables.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
Subtraction between normal variables:
When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.
1999:
Of 100, 20 were in the bottom thid. So
[tex]p_B = \frac{20}{100} = 0.2[/tex]
[tex]s_B = \sqrt{\frac{0.2*0.8}{100}} = 0.04[/tex]
2001:
Of 100, 10 were in the bottom third, so:
[tex]p_A = \frac{10}{100} = 0.1[/tex]
[tex]s_A = \sqrt{\frac{0.1*0.9}{100}} = 0.03[/tex]
To determine this, you test the hypotheses H0 : p1 = p2 , Ha : p1 > p2.
Can also be rewritten as:
[tex]H_0: p_B - p_A = 0[/tex]
[tex]H_1: p_B - p_A > 0[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{s}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, and s is the standard error.
0 is tested at the null hypothesis:
This means that [tex]\mu = 0[/tex]
From the sample:
[tex]X = p_B - p_A = 0.2 - 0.1 = 0.1[/tex]
[tex]s_A = \sqrt{s_A^2+s_B^2} = \sqrt{0.03^2+0.04^2} = 0.05[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{s}[/tex]
[tex]z = \frac{0.1 - 0}{0.05}[/tex]
[tex]z = 2[/tex]
P-value of the test and decision:
The p-value of the test is the probability of finding a difference of proportions of at least 0.1, which is 1 subtracted by the p-value of z = 2.
Looking at the z-table, z = 2 has a p-value of 0.976.
1 - 0.976 = 0.024, so the p-value is given by option B.
The p-value of the test is 0.024 < 0.05(standard significance level), which means that there is enough evidence that the proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced.
